How do I verify the trigonometric identity?

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How do I verify the trigonometric identity?

Verify the trigonometric identity:
$sec x (csc x - 2 sin x) = cot x - tan x$ $secx(cscx-2sinx)=cotx-tanx$

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Answer 1 · 2018-04-01T23:59:59

Using the identities:
$\frac{1}{cos x} = sec x$ $1/cosx= secx$
$\frac{1}{sin x} = csc x$ $1/sinx= cscx$
$1 - {sin}^{2} x = {cos}^{2} x$ $1-sin^2x= cos^2x$
$\frac{sin x}{cos x} = tan x$ $sinx/cosx= tanx$
$\frac{cos x}{sin x} = cot x$ $cosx/sinx= cotx$

Start:
$sec x (csc x - 2 sin x) = cot x - tan x$ $secx(cscx-2sinx)= cotx-tanx$

$\frac{1}{cos x} (\frac{1}{sin x} - 2 sin x) = cot x - tan x$ $1/cosx(1/sinx-2sinx)=cotx-tanx$

$\frac{1}{sin x cos x} - \frac{2 sin x}{cos x} = cot x - tan x$ $1/(sinxcosx)-(2sinx)/cosx=cotx-tanx$

$\frac{1}{sin x cos x} - \frac{2 {sin}^{2} x}{cos x sin x} = cot x - tan x$ $1/(sinxcosx)-(2sin^2x)/(cosxsinx)=cotx-tanx$

$\frac{1 - 2 {sin}^{2} x}{cos x sin x} = cot x - tan x$ $(1-2sin^2x)/(cosxsinx)=cotx-tanx$

$\frac{1 - {sin}^{2} x - {sin}^{2} x}{cos x sin x} = cot x - tan x$ $(1-sin^2x-sin^2x)/(cosxsinx)=cotx-tanx$

$\frac{{cos}^{2} x - {sin}^{2} x}{cos x sin x} = cot x - tan x$ $(cos^2x-sin^2x)/(cosxsinx)=cotx-tanx$

$\frac{{cos}^{2} x}{cos x sin x} - \frac{{sin}^{2} x}{cos x sin x} = cot x - tan x$ $(cos^2x)/(cosxsinx)-(sin^2x)/(cosxsinx)=cotx-tanx$

$\frac{cos x}{sin x} - \frac{sin x}{cos x} = cot x - tan x$ $cosx/sinx-sinx/cosx=cotx-tanx$

$cot x - tan x = cot x - tan x$ $cotx-tanx= cotx-tanx$

Answer 2 · 2018-04-02T00:00:11

See below.

Explanation:

When verifying such identities, it's best to leave one side alone, and begin working with the more complex side. In this case, that would be the left side.

So, recall that $sec x = \frac{1}{cos x}, csc x = \frac{1}{sin x} :$ $secx=1/cosx, cscx=1/sinx:$

$\frac{1}{cos x} (\frac{1}{sin x} - 2 sin x) = cot x - tan x$ $1/cosx(1/sinx-2sinx)=cotx-tanx$

Distribute:

$\frac{1}{cos x sin x} - \frac{2 sin x}{cos x} = cot x - tan x$ $1/(cosxsinx)-(2sinx)/cosx=cotx-tanx$

Subtract the two expressions on the left side, using $cos x sin x$ $cosxsinx$ as a common denominator.
$\frac{1 - 2 {sin}^{2} x}{cos x sin x} = cot x - tan x$ $(1-2sin^2x)/(cosxsinx)=cotx-tanx$

Recall the identity $cos 2 x = 1 - 2 {sin}^{2} x = {cos}^{2} x - {sin}^{2} x .$ $cos2x=1-2sin^2x=cos^2x-sin^2x.$ Apply to the numerator:

$\frac{{cos}^{2} x - {sin}^{2} x}{cos x sin x} = cot x - tan x$ $(cos^2x-sin^2x)/(cosxsinx)=cotx-tanx$

Now, split up the fraction:

$\frac{{cos}^{2} x}{cos x sin x} - \frac{{sin}^{2} x}{cos x sin x} = cot x - tan x$ $cos^2x/(cosxsinx)-sin^2x/(cosxsinx)=cotx-tanx$

Simplify by cancelling out relevant terms:

$\frac{cos x}{sin x} - \frac{sin x}{cos x} = cot x - tan x$ $cosx/sinx-sinx/cosx=cotx-tanx$

Recalling that $cot x = \frac{cos x}{sin x}, tan x = \frac{sin x}{cos x},$ $cotx=cosx/sinx, tanx=sinx/cosx,$ this holds:

$cot x - tan x = cot x - tan x$ $cotx-tanx=cotx-tanx$

Answer 3 · 2018-04-02T00:03:53

Use the definition of the trig functions to rewrite the problem:

$sec x = \frac{1}{cos x}$ $secx=1/cosx$

$csc x = \frac{1}{sin x}$ $cscx=1/sinx$

$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$

$cot x = \frac{1}{tan x} = \frac{1}{\frac{sin x}{cos x}} = \frac{cos x}{sin x}$ $cotx=1/tanx=1/(sinx/cosx)=cosx/sinx$

and also the Pythagorean identity:

${cos}^{2} + {sin}^{2} x = 1$ $cos^2+sin^2x=1$

$\Rightarrow 1 - {sin}^{2} x = {cos}^{2} x$ $=>1-sin^2x=cos^2x$

Now, rewrite the problem in terms of sine and cosine. I'll start with the left side and manipulate it until it looks exactly like the right side:

$L H S = sec x (csc x - 2 sin x)$ $LHS=color(red)secx(color(blue)cscx-2sinx)$

$L H S = \frac{1}{cos x} (\frac{1}{sin x} - 2 sin x)$ $color(white)(LHS)=color(red)(1/cosx)(color(blue)(1/sinx)-2sinx)$

$L H S = \frac{1}{cos x} \cdot \frac{1}{sin x} - \frac{1}{cos x} \cdot 2 sin x$ $color(white)(LHS)=color(red)(1/cosx)*color(blue)(1/sinx)-color(red)(1/cosx)*2sinx$

$L H S = \frac{1}{cos x sin x} - \frac{1}{cos x} \cdot 2 sin x$ $color(white)(LHS)=1/(cosxsinx)-color(red)(1/cosx)*2sinx$

$L H S = \frac{1}{cos x sin x} - \frac{2 sin x}{cos x}$ $color(white)(LHS)=1/(cosxsinx)-(2sinx)/cosx$

$L H S = \frac{1}{cos x sin x} - 2 \cdot \frac{sin x}{cos x}$ $color(white)(LHS)=1/(cosxsinx)-2*sinx/cosx$

$L H S = \frac{1}{cos x sin x} - 2 \cdot tan x$ $color(white)(LHS)=1/(cosxsinx)-2*tanx$

$L H S = \frac{1}{cos x sin x} - tan x - tan x$ $color(white)(LHS)=1/(cosxsinx)-tanx-tanx$

$L H S = \frac{1}{cos x sin x} - \frac{sin x}{cos x} - tan x$ $color(white)(LHS)=1/(cosxsinx)-sinx/cosx-tanx$

$L H S = \frac{1}{cos x sin x} - \frac{sin x \cdot sin x}{cos x \cdot sin x} - tan x$ $color(white)(LHS)=1/(cosxsinx)-(sinxcolor(red)(*sinx))/(cosxcolor(red)(*sinx))-tanx$

$L H S = \frac{1}{cos x sin x} - \frac{{sin}^{2} x}{cos x sin x} - tan x$ $color(white)(LHS)=1/(cosxsinx)-sin^2x/(cosxsinx)-tanx$

$L H S = \frac{1 - {sin}^{2} x}{cos x sin x} - tan x$ $color(white)(LHS)=(1-sin^2x)/(cosxsinx)-tanx$

$L H S = \frac{{cos}^{2} x}{cos x sin x} - tan x$ $color(white)(LHS)=cos^2x/(cosxsinx)-tanx$

$color(white)(LHS)=cos^color(red)cancelcolor(black)2x/(color(red)cancelcolor(black)cosxsinx)-tanx$

$color(white)(LHS)=cosx/sinx-tanx$

$color(white)(LHS)=cotx-tanx$

$color(white)(LHS)=RHS$

The identity is proved. Hope this helped!

Answer 4 · 2018-04-04T14:06:24

$LHS=secx(csc-2sinx)$

$=1/cosx(1/sinx-2sinx)$

$=1/(cosxsinx)-(2sinx)/cosx$

$=(cos^2x+sin^2x)/(cosxsinx)-2tanx$

$=cos^2x/(cosxsinx)+sin^2x/(cosxsinx)-2tanx$

$=cosx/sinx+sinx/cosx-2tanx$

$=cotx+tanx-2tanx$

$=cotx-tanx=RHS$

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How do I verify the trigonometric identity?

Verify the trigonometric identity:
$sec x (csc x - 2 sin x) = cot x - tan x$ $secx(cscx-2sinx)=cotx-tanx$

4 Answers

Explanation:

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